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If \( 1, w, w^{2} \) are three cube roots of unity, then \( \left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right) \) is
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\( 04 \)
We have, $\left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right) \rightarrow(1)$
We know that, $1+\omega+\omega^{2}=0$
$\Rightarrow 1-\omega+\omega^{2}=-2 \omega\{$ subtracting $2 \omega$ both the sides $\}$
$\Rightarrow 1+\omega-\omega^{2}=-2 \omega^{2}\left\{\right.$ subtracting $2 \omega^{2}$ both the sides $\}$
So, Eq. (1) becomes
$=(-2 \omega)\left(-2 \omega^{2}\right)=4 \omega^{3}=4\left\{\right.$ Since,$\left.\omega^{3}=1\right\}$
We know that, $1+\omega+\omega^{2}=0$
$\Rightarrow 1-\omega+\omega^{2}=-2 \omega\{$ subtracting $2 \omega$ both the sides $\}$
$\Rightarrow 1+\omega-\omega^{2}=-2 \omega^{2}\left\{\right.$ subtracting $2 \omega^{2}$ both the sides $\}$
So, Eq. (1) becomes
$=(-2 \omega)\left(-2 \omega^{2}\right)=4 \omega^{3}=4\left\{\right.$ Since,$\left.\omega^{3}=1\right\}$
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